$\lim_{x\to\infty}\dfrac{3x^2}{e^x}=?$ Choose 1 answer: Choose 1 answer: (Choice A) A $0$ (Choice B) B $3$ (Choice C) C $6$ (Choice D) D $\infty$
Solution: $\lim_{x\to\infty} 3x^2=\infty$ and $\lim_{x\to\infty} e^x=\infty$, so $\lim_{x\to\infty}\dfrac{3x^2}{e^x}$ results in the indeterminate form $\dfrac{\infty}{\infty}$. We should use l'Hôpital's rule. $\begin{aligned} &\phantom{=}\lim_{x\to\infty}\dfrac{3x^2}{e^x} \\\\ &=\lim_{x\to\infty}\dfrac{\dfrac{d}{dx}\left[3x^2\right]}{\dfrac{d}{dx}[e^{x}]} \gray{\text{l'Hôpital's rule}} \\\\ &=\lim_{x\to\infty}\dfrac{6x}{e^x} \\\\ &=\lim_{x\to\infty}\dfrac{\dfrac{d}{dx}\left[6x\right]}{\dfrac{d}{dx}[e^{x}]} \gray{\text{l'Hôpital's rule}} \\\\ &=\lim_{x\to\infty}\dfrac{6}{e^x} \\\\ &=0 \end{aligned}$ Note that we used l'Hôpital's rule twice, because the first time we used it, we ended with the indeterminate form $\dfrac{\infty}{\infty}$ too. Also note that we were only able to use l'Hôpital's rule because the limit $\lim_{x\to\infty}\dfrac{\dfrac{d}{dx}\left[6x\right]}{\dfrac{d}{dx}[e^{x}]}$ can actually be determined. In conclusion, $\lim_{x\to\infty}\dfrac{3x^2}{e^x}=0$.